Find Large Inductance Without Meter

Nov 24, 2016 49 Replies

Dream on....if inductor's internal resistance is too big, there will be *NO* L/R curve; one will see a nice resistive divider.

Been there.

200k or so should be enough series resistance.

Wouldn't the other way be to put it into a circuit with an active negative resistive component to make an oscillator.

Or choose a C such that wL >> R and any basic LC oscillator circuit.

Regards, Martin Brown

The application of this device will probably depend to a large extent on it's ability to reject 60Hz, or common power frequencies. 10Hz is just low enough to work around the issue.

So picking a cap that resonates in parallel with the coil at 60 Hz is probably useful, in that, when placed in series, it could aid in rejection.

Establishing 'L' with the random known cap value should allow you to aim for appropriate C values. A simple circuit to model....

RL

If the internal resistance is too big *compared to its inductance*, you are correct. How much inductance must it have to show a L/R curve if it has 20k resistance?

If the inductance is 2000Hy, it will take about .1 sec to reach .63mA if

20V is applied. You might argue that 2000Hy is unrealistically large, but that is what we are trying to learn, isn't it?

This reminds me of a story, about an oral quiz of a physics degree candidate, who was asked how he would measure an inductor.

He replied, 'Connect a battery, switch it through an ammeter, and note the current and time'

"Oh, but wouldn't you use a function generator..." inquired the examiner.

"No, just the battery and a switch. I've done it many times. You take about four turns around a battleship..."

Winfield's suggestion is best. No need to look for the time constant. di/dt does it all.

Since the OP apparently wants to measure the inductance in order to calculate the size of capacitor necessary to have the LC tank resonate at

10 Hz, why doesn't he just throw random capacitors in parallel until it resonates at 10 Hz? Skip the middleman and all that...

capacitors are easily had with capacitance over an range of 12 orders of magnitude, where to start?

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I don't think it would be a good idea to start at 1pF and work up. It shouldn't be too hard to eliminate a few of those orders without testing.

One other means not mentioned much these days is the so called "Three Voltmeter" method. You will need a resistor in series with the coil and signal source. Then measure the ac voltage across the coil, the resistor, the source and do some trigonometry.

Three voltmeter method was used in the olden days and was considered as being well suited to measuring inductance and loss resistance of large lossy iron-cored inductors.

piglet

What happens that no resonance is found due to low Q? A lot of time wasted there.

Then the inductor isn't going to be of much use to him, is it?

It might be if he has a moving magnetic field nearby, if you can think that reasonably.

< snip >

This is how I'd do it, but I'd measure the time constant on charge. Put a resistor or diode across whatever you use to switch the voltage source out of the circuit, to contain the kickback.

Tim Wescott Control systems, embedded software and circuit design I'm looking for work! See my website if you're interested http://www.wescottdesign.com

It will work either in charge or discharge. However, I suggested discharge because I didn't want to run into a problem with current limiting in the power supply which will increase the charge time. However, I just noticed that the coil is wound on a mu-metal core, which delivers very high relative permeability, but saturates at fairly low currents, so that's probably not a real consideration.

Also, I'm not quite sure if this is a problem, but my guess(tm) is this magic "earth sensing magnetic coil" was intended to be a magnetic field sensor, not an electromagnet with current applied. As a sensor, back-EMF is not a problem. However, as an electromagnet, the high voltage produced could easily arc between turns, burn off some insulation, and ruin the magic coil. Methinks that one of the other methods will potentially be less destructive.

Jeff Liebermann jeffl@cruzio.com 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558

Set up your function generator output in series with a high resistor value.

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gnjihj.jpg Drive the inductance at various frequencies and measure the voltage across the inductance. Calculate the impedance using the voltage divider equation , and plot as shown here:
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Now curve-fit the impedance of a known inductance, and tweak the inductance until the two curves match. In the case shown here, the inductance varied with frequency, since an iron core was used.

The 20V (or whatever) power supply will limit at 1mA (or whatever)?

Thanks. That's another approach I will definitely try.

Kevin Foster

lue.

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oss the inductance. Calculate the impedance using the voltage divider equa tion, and plot as shown here:

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ance until the two curves match. In the case shown here, the inductance var ied with frequency, since an iron core was used.

For best results, make Vout < 0.2Vin. This is how I record the data in Mat hcad:

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This is the math:

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This is a nice approach, since it will show you the resistive part, the ind uctive part, the LC resonance, and the capacitive part, if you sweep the fr equency high enough. But remember that your probe and circuit C will be in parallel with the inductance being measured. So you should minimize them.

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